The unit we been use for solubility in this experiment is gram/100-mL. There are many kinds of units to describe how much solute we have in a solvent. There are some examples of other unit that describes solubility. That is g/mol, g/litre, mol/litre, mass percent, mass per volume, and parts per million.
Below the table displays the solubility of some common substances in g/100 mL at the specified temperature.
Solubility process is a process of solubilization for organic solute in water in a mechanistic perspective. First, it involves the break up of solute-solute intermolecular bonds. And then followed by the break up of solvent-solvent intermolecular bonds. After the solutes and solvents intermolecular bonds are break up, the formation of cavity in solvent phase is appearing to be large enough to accommodate solute molecule. The solute will vaporize into the cavity of solvent phase. And the formation of solute-solvent intermolecular bonds will appear. While then the solvent-solvent bonds will reform with solvent restructuring.
Every in our life solubility do play in its important roles. For examples, the salt we use for every day meals. The solubility of the salt will outcome whether the food is tasty if the salt is solute proper into the soup or foods or too sour if the salt is putting too much where it can’t be dissolve into the soup or foods. Coffee powder and sugar is another common examples of solubility in our day life. When having a cup of coffee, for old folks, they prefer coffee that is more saturated or sometimes supersaturated, they will dissolve lesser sugar with more coffee powder into a single cup. But if for youngster, they feel that coffee is tasty only if it is sweet. So, more sugar less coffee powder is a choice for them. And when the sugar are can’t be dissolve in water, they will add hot water instead of cold water because of the influences of high temperature that incite the solubility.
Apparatus and Materials:
-
KClO3 salt
- Distilled water
- Pestle and mortar
-
Measuring cylinder/ Graduated cylinder (100cm3)
-
Beaker (500cm3)
- Porcelain dish
- Thermometer
- Analytical Balance
- Oven
- Pipette (>5-mL)
Procedures:
-
About 2g of KClO3 salt are grind in a mortar by using a pestle.
-
0.5g of the fine salt is transfer to a measuring cylinder. About 10cm3 of distilled water is added to the measuring cylinder.
-
While adding KClO3 stirring is provided a little at a time until no more salt can dissolve where is until a saturated solution is obtained and the solubility of the salt is maximum.
-
Measuring cylinder is placed in hot water in a beaker. (It is best with the temperature of the water is 2oC higher than the required temperature in determining the solubility temperature. For examples, if the required solubility temperature is 40 oC, the temperature of the hot water should be 42 oC.)
-
The solution is stirred and if all the salt dissolved, more KClO3 salt is added while stirring. Process is continuing until a saturated solution is obtained.
-
5cm3 of the solution is transferred into a crucible using the pipette that has already been accurately weighed.
-
The salt solution in the crucible is allowed to evaporate on a hot water bath until dry. The salt solution is further dried in an oven at 110oC until a constant weight is obtained.
- The final mass is recorded in a proper table.
-
The above procedure for the determination of solubility of KClO3 salt is repeated at various temperatures for examples: 40 oC, 50 oC, 60 oC, and 80 oC.
- A graph of solubility against temperature is plotted.
Data and Result:
Calculation and Analysis:
The Solubility of KClO3 salt in 40 oC
= 0.36g/5mL
To convert into g/100mL,
0.36g x 20
= --------------
5mL x 20
= 7.20g/100mL
The Solubility of KClO3 salt in 50 oC
= 0.48g/5mL
To convert into g/100mL,
0.48g x 20
= --------------
5mL x 20
= 9.60g/100mL
The Solubility of KClO3 salt in 50 oC
= 0.87g/5mL
To convert into g/100mL,
0.87g x 20
= --------------
5mL x 20
= 17.40g/100mL
The Solubility of KClO3 salt in 50 oC
= 1.50g/5mL
To convert into g/100mL,
1.50g x 20
= --------------
5mL x 20
= 30.00g/100mL
Discussion:
In this experiment, we try to determine the solubility of KClO3 in different temperature (oC). In order to find the solubility, at starting of the experiment we have to grind finely about 2g of KClO3 salt in a mortar. We need to grind the KClO3 salt is to increase the surface area of KClO3 salt so that it will dissolve easily in distilled water. After that we transfer carefully the 5g grinded KClO3 salt into a measuring cylinder and add 10 cm of distilled water as well. When u transfers the grinded KClO3 salt into measuring cylinder, you have to be careful because u might drop some KClO3 salt at the top of the measuring cylinder. This will waste your KClO3 salt and to retrieve the salt which left at the top of the measuring cylinder, you have to move some distilled water from the bottom of the measuring cylinder to top of the measuring cylinder. This might be able to help you to retrieve the KClO3 salt which left on the top of the measuring cylinder.
During experiment, we add the grinded KClO3 salt a little at a time and stir it until no more can dissolve in the distilled water. This means the grinded KClO3 salt present is in excess. We stir the solution while adding the grinded KClO3 salt is because we need to accelerate the speed of grinded KClO3 salt dissolve and can dissolve in balance. If there is no more KClO3 salt can be dissolve in that distilled water means that you have obtain solution that considered as a saturated solution and the solubility of the salt is maximum.
Later on, the measuring cylinder is placed in a beaker that contain of hot water. At the beginning of the experiment, we use 40 oC as the temperature of the hot water in the beaker. But in this situation it is best if the temperature of the water is 2 oC higher than the required temperature in determining the solubility temperature. Therefore the temperature is required of the hot water should be 42 oC. We need to make sure the temperature is 2 oC higher than the required temperature because there is temperature gradient between the test tube and the water inside the beaker. Besides this, there is no 100% transfer of heat from water bath to the KClO3 solution inside the test tube. Some heat will be transferred to the environment. Heat loss will also occur due to convection. To ensure the temperature in the test tube is 40 oC so we need to make sure the temperature is 2 oC higher than the required temperature.
Up next, we stir solution to ensure that we have obtain a solution that can considered as saturated solution and if all the grinded KClO3 salt dissolves, add more grinded KClO3 salt into the solution. While u adds the grinded KClO3 salt, you have to add grinded KClO3 salt in gradually and cannot add too many in once. If the solution is saturated, you might make the solution contain too many grinded KClO3 salt and will make u difficult to transfer the saturated solution into crucible for the next step. If haven’t obtain a saturated solution, thus continue add more grinded KClO3 salt until a saturated solution is obtained. By the way, we also have to control the temperature during the process of heating so that we can determine the solubility of KClO3 at a certain temperature.
After we have successfully obtained a solution that can considered as saturated solution, we can transfer the 5 cm of the saturated solution by using pipette (5ml) into a crucible which has already been accurately weighed. When we transfer 5 cm of saturated solution by using 5ml pipette into crucible, we have to be careful and must transfer only the top of the saturated solution. If we accidentally pipetted the KClO3 salt in the bottom of the measuring cylinder, we have pipette all over again. This is because the KClO3 salt in the bottom of the measuring cylinder is not a part of the saturated solution. If we didn't pipette all over again, the result you obtain is can consider not accurate. Other than that, it is best way if u pipette once for 1 cm and pipette 5 times for 5 cm. This is because if you pipette 5 cm in once and cannot transfer into crucible in time, the KClO3 solution will freeze and back into powder form. If the KClO3 solution freeze and back into powder form, it will stuck at the exit of the pipette and that the solution cannot go out from the pipette. Therefore you cannot obtain the solution from what you have pipetted.
Next step of the experiment is put the crucible on a hot water bath. The purpose of putting the crucible on a hot water bath is let the KClO3 solution to evaporate until dry. We can notice the KClO3 solution will slowly become crystallize forms. If we noticed this happen, that means we can put into the oven at 110 oC for certain to further dries and fully evaporate until constant weight is obtained. After you notice the KClO3 crystal has fully evaporate and completely dry, put the KClO3 crystal into the desiccators. The purpose of putting the KClO3 crystal into the desiccators is let the KClO3 crystal completely dry in a clean place and place without foreign substance. The desiccators gel silica able to absorb the water that still contain inside the desiccators. After about 10mins, the KClO3 crystal can weigh accurately using electrical balance.
In this experiment, we also try to obtain the solubility of KClO3 salt in different temperature and it is 50 oC, 60 oC and 80 oC. In order to find the solubility of KClO3 salt, all the procedure above must be repeated. In obtaining the solubility of KClO3 salt in 80 oC must be more careful, because it is very hot. Beside that, to transfer the KClO3 solution into crucible must be done it very fast and carefully because the freezing rate is faster than 40 oC.
Finally, the solubility of KClO3 salt in 40 oC, 50 oC, 60 oC and 80 oC had obtained. So a graph of solubility of salt vs. temperature must be plot according the solubility of KClO3 salt which had obtained. For 40 oC, the constant actual salt mass is 0.36g. For 50 oC, the constant actual salt mass is 0.48g. For 60 oC, the constant actual salt mass is 0.87g. While for 80 oC, the constant actual salt mass is 1.50g. After the calculation and conversion into the solubility of salt, the solubility of KClO3 salt in 40 oC is 7.20g/100mL. Meanwhile, the solubility of KClO3 salt in 50 oC is 9.60g/100mL and the solubility of KClO3 salt in 60 oC is 17.40g/100mL. Lastly, the solubility of KClO3 salt in 80 oC is 30.00g/100mL. As we can see, the solubility of KClO3 salt in different temperature is respectively increasing when the temperature is higher. So we know the higher of the temperature of the solvent (distilled water), the more solute (KClO3 salt) that will dissolve in it.
Solubility is sensitive to temperature. In order for a solvent to dissolve a solute, the particles of the solvent must be able to separate the particles of the solute and occupy the intervening spaces. Polar solvent molecules can effectively separate the molecules of other polar substances. This happens when the positive end of a solvent molecule approaches the negative end of a solute molecule. A force of attraction then exists between the two molecules. The solute molecule is pulled into solution when the force overcomes the attractive force between the solute molecule and its neighboring solute molecule. In this situation, the main factor is temperature. So the heat that provide will help to separate the particles of the solute (KClO3 salt) and occupy the intervening spaces.
Precaution steps:
- We have considered the temperature, because a constant temperature must be maintain when obtaining the solubility of certain salt solution.
- Using the pipette is very critical in this experiment. We have to transfer the solution as fast as possible because it might freeze and stacked at the exit of the pipette.
-
To ensure the KClO3 crystal has fully evaporate and completely dry, we must put the KClO3 crystal into the oven for about 110 oC and put it a bit longer for about 15mins.
-
Before weigh the KClO3 crystal by electrical balance, we have to put into desiccators.
-
A constant actual salt mass have to obtain to make sure that all the water inside is evaporated completely.
Conclusion:
-
The solubility of KClO3 solution in 40 oC is 7.20g/100 ml.
-
The solubility of KClO3 solution in 50 oC is 9.60g/100ml.
-
The solubility of KClO3 solution in 60 oC is 17.40g/100ml.
-
The solubility of KClO3 solution in 80 oC is 30.00g/100ml.
- Solubility is sensitive to temperature.
- Generally, an increase in the temperature of the solution increases the solubility of a solid solute. A few solid solutes, however, are less soluble in warmer solutions. For all gases, solubility decreases as the temperature of the solution rises.
References:
- http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch18/soluble.html
- http://members.tripod.com/~EppE/soluble.htm
- http://educ.queensu.ca/~science/main/concept/chem/c10/C10lalg2.htm
- http://chemistry.about.com/library/weekly/blsolubility.htm
- Chemistry fourth Edition, Silberberg